Dividing an unit interval ([0,1]) into two equal subintervals by the midpoint (dfrac {0+1} {2}=dfrac {1} {2}), denote the left subinterval by (I_{1}=left[ 0,dfrac {1} {2^{1}}
ight]), next, divide (I_{1}) into two equal parts by its midpoint, denote the left subinterval by (I_{2}=left[ 0,dfrac {1} {2^{2}}
ight]). Keep repeating this procedure indefinitely, what's left in the end ? Since referred infinitely many times here, it seems impossible to image the end case, but we could actually 'see' it!
Continue the process, obtaining a sequence of nested intervals$$I_{n}=left[ 0,dfrac {1} {2^{n}}
ight], n = 1, 2, 3, ... $$Applying the nested intervals theorem there is only one point, one real number 0 contained in every (I_{n}), i.e. $$displaystyleigcap_{{n=1}}^{infty}left[ 0,dfrac {1} {2^{n}}
ight]=left[ 0,0
ight]={0}$$
In conclusion, the only thing left after infinitely many times of these dividing is a point. More general, we don't need to divide each interval equally to form that sequence of nested intervals, since the nested intervals theorem doesn't requires that.